Is there a dense set in $\mathbb{R}$ with outer measure $1$? And what about with the Density Topology?

Question

I'm looking for a dense set in $\mathbb{R}$ with outer measure $1$. There is an example like that, but in $[0,1]$ Vitali set of outer-measure exactly $1$.. Also I've tried with Bernstein sets but I don't get it.

Any hint is appreciated.

Edit: I've been trying to understand the affirmation that makes F.D Tall in his articule: https://msp.org/pjm/1976/62-1/pjm-v62-n1-p25-p.pdf In the proof of theorem 4.13, he said:

...A set of outer measure $1$ is clearly dense...

In the Density Topology, $\mathbb{Q}$ is not dense. In fact, any countable subset is not dense, moreover, any nullset is closed and discrete, and viceversa. But this topology contains the euclidean topology, so if you have interesting dense subsets in $\mathbb{R}$ with outer measure $1$, I would appreciate it.

Answer 1

In the usual (Euclidean) topology, $\Bbb R$ has a dense open subset $S$ with measure $m(S)=1.$

Let $\Bbb Q=\{q_i: i\in \Bbb N\}.$ Let $K(1)$ be an open interval with irrational end-points, with $m(K(1))=2^{-1}$ and $q_1\in K(1).$

For $n\in \Bbb N,$ suppose that $K(n)$ is a union of finitely many pair-wise disjoint bounded open intervals with irrational end-points, and that $m(K(n))=1-2^{-n}.$ Construct $K(n+1)$ as follows:

$K(n)$ is bounded so $\Bbb Q \not \subset K(n).$ Let $i(n)$ be the least $i\in \Bbb N$ such that $q_i\not \in K(n).$ Since $q_{i(n)} \not \in \overline {K(n)},$ let $U(n)$ be an open interval with irrational end-points, with $0<m(U(n))<2^{-(n+1)},$ and with $q_{i(n)}\in U(n)\subset \Bbb R \setminus K(n).$ Since $K(n)\cup U(n)$ is bounded, let $ V(n)$ be an open interval with irrational end-points, with $V(n)$ disjoint from $K(n)\cup U(n)$ and with $ m(V(n))=2^{-(n+1)}-m(U(n)).$

Let $K(n+1)=K(n)\cup U(n)\cup V(n).$

Let $S=\cup_{n\in \Bbb N}K(n).$

It should be clear that $S$ is open and $m(S)=1.$ We show that every $q_n\in S$ by induction on $n,$ as follows:

We have $q_1\in S.$ For $n\in \Bbb N,$ if $\{q_i:i\le n\}\subset S,$ then for each $ i\le n$ let $f(i)$ be the least (or any) $j$ such that $q_i\in K(j),$ and let $g(n)=\max \{f(i):i\le n\}.$

Now if $q_{n+1}\in K(g(n))$ then $q_{n+1}\in S.$

But if $q_{n+1}\not \in K(g(n)),$ then, since $\{q_i:i\le n\}\subset \cup_{i\le n}K(f(i))=K(g(n)),$ we have, from the recursive construction of $K(g(n)+1)$ from $K(g(n)),$ that $i(g(n))=n+1,$ so $q_{n+1}\in U(g(n))\subset K(g(n)+1)\subset S.$

Answer 2

Take $\mathbb{Q}\cup[0,1]$, for instance.

Answer 3

Let $\{a_1, a_2, \cdots, a_n, \cdots\}$ be the set of all rational numbers. For each $i$, take a neighborhood of length $\frac{1}{2^{i+1}}$ near $a_i$ and denote it by $B_i$. Let $U=\cup B_i$. Then $|U| \le \Sigma |B_i| \le \frac{1}{2}$. Theore, $U$ is an open dense subset of measure less than $1$. Scale it to any measure you want.

If you need a non-measurable set, just take the union of this set and the Vatali set in the unit interval.